Question

Find the deff equation $(x^2+y^2)dx-2xydy=0$

Answer 1

$(x^2+y^2)dx-2xydy=0$

$Mdx-Ndy=0$

$M_y\frac{\partial }{\partial y}(x^2+y^2)=2y$

$N_x\frac{\partial N}{\partial x}=\frac{\partial }{\partial x}(-2xy)=-2y$

equation is not exact as $\Rightarrow M_y\ne N_x$

calculating the value for:

$\frac{(M_y.N_x)}{N}=\frac{|2y|-|-2y|}{2xy}=\frac{4y}{2xy}=\frac{-2}{x}$

$\frac{(M_y-N_x)}{N}=\frac{-2}{x}\Rightarrow f\left(x\right)\frac{-2}{x}$

As $f\left(x\right)$ can be considered as a function of $x$ only, the integrating factor denoted by $I$ is given by $I=e^{\int f\left(x\right)dx}=e^{\int -\frac{-2}{x}dx}\Rightarrow I=e^{-2\ln x}=x^{-2}$

Multiplying the integrating factor I on both sides

$(1+x^{-2}{y}^2)dx-2x^{-2}ydy=0$

equation is in the form of $Mdx+Ndy=0$

$M_y=N_x=2x^{2}y$

$\int M.\partial x=\int (1+x^{-2}{y}^2{)}^{\partial x}=1+x^{-2}{y}^2/-1$

$\int M.\partial x=x-x^{-1}{y}^2$

$\int M\partial y=\int -2x^{-3}y\partial y=\frac{-2x^{-1}{y}^2}{2}\Rightarrow x-x^{-1}{y}^2=C$